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\begin{document}

\title{Complex Analysis}
\subtitle{Chapter 6. Conformal Mapping, Dirichlet's Problem}
%\institute{SLUC}
\author{LVA}
%\date
%\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
%\date{ {2023年9月21日} }

\maketitle

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\begin{frame}{Contents 1-2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item  The Riemann Mapping Theorem
\begin{enumerate}
\item[1.1.] Statement and Proof
\item[1.2.] Boundary Behavior
\item[1.3.] Use of the Reflection Principle
\item[1.4.] Analytic Arcs
\end{enumerate}

\item  Conformal Mapping of Polygons
\begin{enumerate}
\item[2.1.] The Behavior at an Angle
\item[2.2.] The Schwarz-Christoffel Formula
\item[2.3.] Mapping on a Rectangle
\item[2.4.] The Triangle Functions of Schwarz
\end{enumerate}

\end{enumerate}

\end{frame}

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\begin{frame}{Contents 3-5}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[3.] A Closer Look at Harmonic Functions
\begin{enumerate}
\item[3.1.] Functions with the Mean-value Property
\item[3.2.] Harnack's Principle
\end{enumerate}
 
\item[4.] The Dirichlet Problem
\begin{enumerate}
\item[4.1.] Subharmonic Functions
\item[4.2.] Solution of Dirichlet's Problem
\end{enumerate}

\item[5.] Canonical Mappings of Multiply Connected Regions
\begin{enumerate}
\item[5.1.] Harmonic Measures
\item[5.2.] Green's Function
\item[5.3.] Parallel Slit Regions
\end{enumerate}

\end{enumerate}

\end{frame}

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\begin{frame}{1.1. Statement and Proof. The Riemann mapping theorem. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Given any simply connected region $\Omega$ which is not the whole plane, and a point $z_0\in \Omega$, there exists a unique analytic function $f(z)$ in $\Omega$, normalized by the conditions $f(z_0) = 0$, $f'(z_0) > 0$, such that $f(z)$ defines a
one-to-one mapping of $\Omega$ onto the disk $|w| < 1$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.1. Statement and Proof. Exercise - 1}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $z_0$ is real and $\Omega$ is symmetric with respect to the real axis, prove by the uniqueness that $f$ satisfies the symmetry relation $f(\bar{z}) = \overline{f(z)}$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.1. Statement and Proof. Exercise - 2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
What is the corresponding conclusion if $\Omega$ is symmetric with respect to the point $z_0$?
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.2. Boundary Behavior. Theorem 2. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Let $f$ be a topological mapping of a region $\Omega$ onto a region $\Omega'$. 
If $\{z_n\}$ or $z(t)$ tends to the boundary of $\Omega$, then $\{f(z_n)\}$ or $f(z(t))$ tends to the boundary of $\Omega'$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.3. Use of the Reflection Principle. Theorem 3. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Suppose that the boundary of a simply connected region $\Omega$ contains a line segment $\gamma$ as a one-sided free boundary arc. 
Then the function $f(z)$ which maps $\Omega$ onto the unit disk can be extended to a function which is analytic and one to one on $\Omega\cup\gamma$. 
The image of $\gamma$ is an arc $\gamma'$ on the unit circle.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.4. Analytic Arcs. Theorem 4. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If the boundary of $\Omega$ contains a free one-sided analytic arc $\gamma$, then the mapping function has an analytic extension to $\Omega\cup\gamma$, and $\gamma$ is mapped on an arc of the unit circle.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2. Conformal Mapping of Polygons. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
When does the mapping problem have an explicit formula?
}

\item  Answer. 
\begin{enumerate}
\item When $\Omega$ is a polygon, the mapping problem has an almost explicit solution. 

\item Indeed, we shall find that the mapping function can be expressed through a formula in which only certain parameters have values that depend on the specific shape of the polygon.
 
\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.1. The Behavior at an Angle. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[1.]  We assume that $\Omega$ is a bounded simply connected region whose boundary is a closed polygonal line without self-intersections. 

\item[2.]  Let the consecutive vertices be $z_1, \cdots, z_n$ in positive cyclic order  (we set $z_{n+1} = z_1$). 

\item[3.]  The angle at $z_k$ is given by the value of $$\mathrm{arg}\, (z_{k-1}-z_k)/(z_{k+1}-z_k)$$ between $0$ and $2\pi$.

\item[4.] We shall denote it by $\alpha_k\pi$, $0 < \alpha_k < 2$. 

\item[5.] It is also convenient to introduce the outer angles 
$$\beta_k\pi=(1-\alpha_k)\pi,\,\, -1 < \beta_k < 1. $$

\end{enumerate}

\end{frame}

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\begin{frame}{2.1. The Behavior at an Angle. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[6.] Observe that $\beta_1 + \cdots + \beta_n = 2$. 

\item[7.] The polygon is convex if and only if all $\beta_k > 0$.

\item[8.]  We know by Theorem 3 that the mapping function $f(z)$ can be extended by continuity to any side of the polygon (that is, to the open line segment between two consecutive vertices), and that each side is mapped in a one-to-one way onto an arc of the unit circle.

\item[9.]  We wish to show that these arcs are disjoint and leave no gap between them.
 
\end{enumerate}

\end{frame}

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\begin{frame}{2.1. The Behavior at an Angle. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[10.]  To see this we consider a circular sector $S_k$ which is the intersection of $\Omega$ with a sufficiently small disk about $z_k$.

\item[11.]  A single-valued branch of $\zeta = (z - z_k)^{1/\alpha_k}$ maps $S_k$ onto a half disk $S_k'$. 

\item[12.]  A suitable branch of $z_k + \zeta^{\alpha_k}$ has its values in $\Omega$, and we may consider the function $g(\zeta) = f(z_k + \zeta^{\alpha_k})$ in $S_k'$. 

\item[13.]  It follows by Theorem 2 that $|g(\zeta)|\to 1$ as $\zeta$ approaches the diameter. 

\item[14.]  The reflection principle applies, and we conclude that $g(\zeta)$ has an analytic continuation to the whole disk.

\end{enumerate}

\end{frame}

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\begin{frame}{2.1. The Behavior at an Angle. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[15.] In particular, this implies that $f(z)$ has a limit $w_k = e^{i\theta_k}$ as $z\to z_k$, and we find that the arcs that correspond to the sides meeting at $z_k$ do indeed have a common end point.

\item[16.] Since $\mathrm{arg}\, f(z)$ must increase as $z$ traces the boundary in positive direction, the arcs do not overlap, at least not in a neighborhood of $w_k$. 

\item[17.] If we take into account that $f$ maps the boundary on a curve with winding number 1 about the origin, it can easily be concluded that all the arcs are mutually disjoint. 

\item[18.] In other words, $f$ can be extended to a homeomorphic map of $\Omega^-$ onto the closed unit disk, the vertices $z_k$ go into points $w_k$, and the sides correspond to the arcs between these points (Fig. 6-1).

\end{enumerate}

\end{frame}

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\begin{frame}{2.1. The Behavior at an Angle. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}


\begin{figure}[ht!]
\centering
\includegraphics[height=0.6\textheight, width=0.75\textwidth]{figure-6-1.png}
%\caption{Fig. 6-1. Conformal Mapping of a polygon }
\end{figure}


\end{frame}

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\begin{frame}{2.2. The Schwarz- Christoffel Formula. Theorem 5. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}

\item The formula we are looking for refers not to the function $f$, but to its inverse function, which we shall denote by $F$.

\item  {\color{red}Theorem 5. 
The functions $z = F(w)$ which map $|w| < 1$ conformally onto polygons with angles $\alpha_k\pi\,(k=1,\cdots,n)$ are of the form
\begin{equation}
F(w) = C\int_0^w \prod\limits_{k=1}^{n}(w-w_k)^{-\beta_k}dw + C'
\label{eq-3}
\end{equation}
where $\beta_k=1-\alpha_k$, the $w_k$ are points on the unit circle, and $C$, $C'$ are complex constants.
}

\end{itemize}

\end{frame}

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\begin{frame}{2.2. The Schwarz- Christoffel Formula. Theorem 5. Proof.}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[1.] Because the function $g(\zeta) = f(z_k + \zeta^{\alpha_k})$ considered in the last paragraph of 2.1 is analytic at the origin, it has a Taylor development
$$
f(z_k + \zeta^{\alpha_k}) = w_k + \sum\limits_{m=1}^{\infty} a_m\zeta^m.
$$

\item[2.] Here $a_1 \neq 0$, for otherwise the image of the half disk $Z_k'$ could not be contained in the unit disk. 

\item[3.] Therefore the series can be inverted, and on setting $w = f(z_k + \zeta^{\alpha_k})$ we obtain
$$
\zeta = \sum\limits_{m=1}^{\infty} b_m(w-w_k)^m
$$
with $b_1 \neq 0$, the development being valid in a neighborhood of $w_k$.

\end{enumerate}

\end{frame}

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\begin{frame}{2.2. The Schwarz- Christoffel Formula. Theorem 5. Proof.}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[4.] We raise to the power $\alpha_k$ and find, in terms of the inverse function $F$,
$$
F(w) - z_k = (w - w_k)^{\alpha_k}G_k(w)
$$
where $G_k$ is analytic and $\neq 0$ near $w_k$. 

\item[5.] It follows by differentiation that $F'(w)(w-w_k)^{\beta_k}$ is analytic and $\neq 0$ at $w_k$, and therefore the product 
\begin{equation}
H(w)=F'(w)\prod\limits_{k=1}^{n} (w-w_k)^{\beta_k}
\label{eq-4}
\end{equation}
is analytic and $\neq 0$ in the closed unit disk. 

\end{enumerate}

\end{frame}

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\begin{frame}{2.2. The Schwarz- Christoffel Formula. Theorem 5. Proof.}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[6.] We claim that $H(w)$ is, in actual fact, a constant.

\item[7.] For this purpose we examine its argument when $w = e^{i\theta}$ lies on the unit circle between $w_k = e^{i\theta_k}$ and $w_{k+1} = e^{i\theta_{k+1}}$.

\item[8.] We know that $\mathrm{arg}\, F'(e^{i\theta})$ equals the angle between the tangent to the unit circle at $e^{i\theta}$ and the tangent to its image at $F(e^{i\theta})$; with an abbreviated notation we express this by 
$$\mathrm{arg}\, F' = \mathrm{arg}\, dF - \mathrm{arg}\, dw.$$

\item[9.] But $\mathrm{arg}\, dF$ is constant because $F$ describes a straight line, and 
$\mathrm{arg}\, dw = \theta + \pi/2$. 

\end{enumerate}

\end{frame}

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\begin{frame}{2.2. The Schwarz- Christoffel Formula. Theorem 5. Proof.}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[10.] The factor $w -w_k$ can be written 
$$e^{i\theta}-e^{i\theta_k} = 2ie^{i(\theta+\theta_k)/2}\sin\frac{1}{2}(\theta-\theta_k),$$ 
and hence its argument is $\theta/2$ plus a constant (this is also evident geometrically).

\item[11.] When we add the arguments of all factors on the right-hand side of (\ref{eq-4}) we find that $\mathrm{arg}\, H(w)$ differs by a constant from 
$-\theta + \left( \sum\limits_{1}^{n}\beta_k\right)\cdot\theta/2=0$.

\item[12.] Thus we conclude that $\mathrm{arg}\, H(w)$ is constant between $w_k$ and $w_{k+l}$,  and since it is continuous it must be constant on the whole unit circle. 

\end{enumerate}

\end{frame}

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\begin{frame}{2.2. The Schwarz- Christoffel Formula. Theorem 5. Proof.}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[13.] The maximum principle permits us to conclude that 
$$\mathrm{arg}\, H(w) = \mathrm{Im}\, \log H(w)$$ 
is constant inside the unit circle, and so is consequently $H(w)$.

\item[14.] We have now proved that 
$$
F'(w) = C \prod\limits_{k=1}^{n} (w-w_k)^{-\beta_k}, 
$$
and formula (\ref{eq-3}) follows by integration.

\end{enumerate}

\end{frame}

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\begin{frame}{2.2. The Schwarz- Christoffel Formula. Theorem 5. Remark.}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[1.] We remark that a linear transformation of the unit circle permits us to place three of the points $w_k$, for instance, $w_1, w_2, w_3$, in prescribed positions.

\item[2.] For $n = 3$ we see that the mapping function depends only on the angles, except for trivial variable transformations; this reflects the fact that triangles with the same angles are similar. 

\item[3.] For $n > 3$ the remaining constants $w_4,\cdots, w_n$, or their arguments $\theta_k$, are called the {\color{blue} accessory parameters} of the problem. 

\item[4.] It is only in rare cases that they can be determined other than by numerical computation.

\end{enumerate}

\end{frame}

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\begin{frame}{2.2. The Schwarz- Christoffel Formula. Theorem 5. Remark.}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[5.] If we give arbitrary values to the $\theta_k$ it is quite easy to verify that a function of the form (\ref{eq-3}) maps the unit circle on a closed polygonal line, but usually we are unable to tell whether it will intersect itself or not. 

\item[6.] If it does not, it is not difficult to show that $F(w)$, as given by (\ref{eq-3}), yields a one-to-one mapping onto the inside of the polygonal line (the precise proof makes use of the argument principle).

\item[7.] Formula (\ref{eq-3}) is known as the {\color{blue}\it Schwarz-Christoffel formula}.

\end{enumerate}

\end{frame}

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\begin{frame}{2.2. The Schwarz- Christoffel Formula. Theorem 5. Remark.}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[8.] Another version of the same formula serves to map the upper half plane onto the inside of a polygon. 

\item[9.] The mapping function, from $\mathrm{Im}\, w > 0$ to $\Omega$, can now be written in the form
\begin{equation}
F(w) = C\int_0^w \prod\limits_{k=1}^{n-1} (w-\xi_k)^{-\beta_k}dw + C'
\label{eq-5}
\end{equation}
where the $\xi_k$ are real. 


\end{enumerate}

\end{frame}

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\begin{frame}{2.2. The Schwarz- Christoffel Formula. Theorem 5. Remark.}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[10.] The last exponent $\beta_n$ does not appear explicitly in the formula, but it is determined by $$\beta_n = 2 - (\beta_1 + \cdots + \beta_{n-1}),$$ and like the other exponents it is subject to the condition $-1 < \beta_n < 1$. 

\item[11.] It then follows that the integral (\ref{eq-5}) converges for $w = \infty$, and the point at $\infty$ will correspond to a vertex with angle $\alpha_n\pi$, $\alpha_n = 1 -\beta_n$. 

\item[12.] If $\beta_n = 0$ the vertex is only apparent, and the polygon reduces to one with $n-1$ sides.


\end{enumerate}

\end{frame}

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\begin{frame}{2.2. The Schwarz- Christoffel Formula. Exercise - 1}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Show that the $\beta_k$ in (3) may be allowed to become $= -1$. 
What is the geometric interpretation?
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.2. The Schwarz- Christoffel Formula. Exercise - 2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If a vertex of the polygon is allowed to be at $\infty$, what modification does the formula undergo? If in this context $\beta_k = 1$, what is the polygon like?
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.2. The Schwarz- Christoffel Formula. Exercise - 3}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Show that the mappings of a disk onto a parallel strip, or onto a half strip with two right angles, can be obtained as special cases of the Schwarz-Christoffel formula.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.2. The Schwarz- Christoffel Formula. Exercise - 4}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Derive formula (5), either directly or with the help of (3).
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.2. The Schwarz- Christoffel Formula. Exercise - 5}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Show that 
$$
F(w)=\int_0^w (1-w^n)^{-2/n}dw
$$
maps $|w| < 1$ onto the interior of a regular polygon with $n$ sides.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.2. The Schwarz- Christoffel Formula. Exercise - 6}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Determine a conformal mapping of the upper half plane on the region $\Omega = \{z = x + iy; x > 0, y > 0, \, \min (x,y) < 1\}$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.3. Mapping on a Rectangle. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[1.] 
In case $\Omega$ is a rectangle we may choose $x_1 = 0, x_2 = 1, x_3 = \rho > 1$ in (\ref{eq-5}). The mapping function will thus be given by
$$
F(w) = \int_0^w \frac{dw}{\sqrt{w(w-1)(w-\rho)}}
$$
which is an {\color{blue}\it elliptic integral}.
 
\item[2.]  
To be unambiguous we decide that the values of $\sqrt{w}$, $\sqrt{w-1}$, and $\sqrt{w-\rho}$ shall lie in the first quadrant. 
%\footnote{Why can we discard the other square root?}. 

\item[3.]  
For a detailed study of the mapping, let us follow $F(w)$ as $w$ traces the real axis. 

\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Mapping on a Rectangle. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[4.]  
When $w$ is real, each of the square roots is either positive or purely imaginary with a positive imaginary part (save for the point where the square root is $0$).

\item[5.]  
As $0 < w < 1$ there are one real and two imaginary square roots. 

\item[6.]  
Therefore $F(w)$ decreases from $0$ to a value $-K$ where
\begin{equation}
K=\int_0^1\frac{dt}{\sqrt{t(1-t)(\rho-t)}}. 
\label{eq-6}
\end{equation}

\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Mapping on a Rectangle. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[7.] 
For $1 < w < \rho$ there is only one imaginary square root. 

\item[8.] 
It follows that the integral from $1$ to $w$ is purely imaginary with a negative imaginary part.

\item[9.] 
Hence $F(w)$ will follow a vertical segment from $-K$ to $-K-iK'$, 
\begin{equation*}
K'=\int_1^\rho\frac{dt}{\sqrt{t(t-1)(\rho-t)}}. 
%\label{eq-6}
\end{equation*}

\item[10.] 
For $w > \rho$ the integrand is positive, and $F(w)$ will trace a horizontal segment in the positive direction. 

\item[11.] 
How far does it extend? 

\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Mapping on a Rectangle. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[12.] 
Since the image is to be a rectangle, it must end at the point $-iK'$, but we prefer a direct verification.

\item[13.] 
One way is to express the length of the segment by the integral
\begin{equation*}
\int_\rho^{\infty}\frac{dt}{\sqrt{t(t-1)(t-\rho)}}. 
%\label{eq-6}
\end{equation*}
and to show by the change of integration variable 
$$t = (\rho - u)/(1 - u)$$
that the integral transforms to (\ref{eq-6}). 

\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Mapping on a Rectangle. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[15.] 
It is easier, however, to observe that Cauchy's theorem yields
\begin{equation*}
\int_{-\infty}^{\infty}\frac{dt}{\sqrt{t(t-1)(t-\rho)}}=0,  
%\label{eq-6}
\end{equation*}
for the integral over a semicircle with radius $R$ tends to $0$ as $R \to \infty$. 

\item[16.] 
The vanishing of the real part implies the equality of the horizontal segments, and from the vanishing of the imaginary part we deduce that $-\infty < w < 0$ is mapped on the segment from $-iK'$ to $0$. 

\item[17.] 
The rectangle is completed.

\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Mapping on a Rectangle. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[18.] 
It is often preferable to use a formula which reflects the double symmetry of the rectangle. 

\item[19.] 
The vertices can be made to correspond to points $\pm 1$ and $\pm 1/k$ with $0<k<1$.

\item[20.] 
Since a constant factor does not matter we can choose the mapping to be given by
\begin{equation}
F(w) = \int_{0}^{w}\frac{dw}{\sqrt{(1-w^2)(1-k^2w^2)}}, 
\label{eq-7}
\end{equation}
and this time we agree that $\sqrt{1-w^2}$ and $\sqrt{1-k^2w^2}$ shall have positive real parts. 

\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Mapping on a Rectangle. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[21.] 
It is seen that the rectangle will have vertices at $-\frac{K}{2}$, $\frac{K}{2}$, $\frac{K}{2}+iK'$, $-\frac{K}{2}+iK'$ where
\begin{equation*}
\begin{aligned}
K &= \int_{-1}^{1}\frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}, \\
K' &= \int_{1}^{1/k}\frac{dt}{\sqrt{(t^2-1)(1-k^2t^2)}}.
\end{aligned}
%\label{eq-6}
\end{equation*}

\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Mapping on a Rectangle. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{figure}[ht!]
\centering
\includegraphics[height=0.6\textheight, width=0.55\textwidth]{figure-6-2.png}
%\caption{Fig. 6-1. Conformal Mapping of a rectangle }
\end{figure}

\end{frame}

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\begin{frame}{2.3. Mapping on a Rectangle. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[22.] 
The image of the upper half plane is the shaded rectangle $R_0$ in Fig.6-2. 

We denote the inverse function of $F$ by $w = f(z)$; it is defined in $R_0$ and can be extended by continuity to a one-to-one mapping of the closed rectangle onto the closed half plane (with the topology of the Riemann sphere).

\item[23.] 
Observe that $z = iK'$ corresponds to $\infty$.

\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Mapping on a Rectangle. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[24.] 
The reflection principle allows us to extend the definition off to the adjacent rectangles $R_1$ and $R_2$, namely by setting $f(z) = \overline{f(\bar{z})}$ for $z\in R_1$ and $f(z) = \overline{f(K-\bar{z})}$ for $z\in R_2$. 

\item[25.] 
Similarly we can pass to $R_3$ either from $R_1$ or $R_2$; the extension is given by $f(z) = f(K - z)$. 

\item[26.] 
The process of reflection can obviously be continued until $f(z)$ is defined as a meromorphic function in the whole plane. 

\item[27.] 
It is perhaps even more convenient to define the extension by periodicity, for we find that the extended function must satisfy 
$$f(z + 2K) = f(z), f(z + 2iK') = f(z).$$

\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Mapping on a Rectangle. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[28.] 
We have shown that the inverse function of the elliptic integral (\ref{eq-7}) is a meromorphic function with periods $2K$ and $2iK'$.

\item[29.] 
Such functions are known as {\color{blue}\it elliptic functions}. 

\item[30.] 
The connection between elliptic integrals and elliptic functions was discovered, but not published, by Gauss; it was rediscovered by Abel and Jacobi.

\end{enumerate}

\end{frame}


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\begin{frame}{2.3. Mapping on a Rectangle. Exercise - 1}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove that formula 
%(7) 
$$
F(w) = \int_0^w \frac{dw}{\sqrt{(1-w^2)(1-k^2w^2)}}
$$
gives $F(\infty) = iK'$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.3. Mapping on a Rectangle. Exercise - 2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Show that $K = K'$ if and only if $k = (\sqrt{2} -1)^2$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.3. Mapping on a Rectangle. Exercise - 3}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Show that $f(z)$, $f(z + K)$, and $f(z + iK')$ are odd functions of $z$ while $f(z + K/2)$ and $f(z + K/2 + iK')$ are even.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.4. The Triangle Functions of Schwarz. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[1.] 
The upper half plane is mapped on a triangle with angles $\alpha_1\pi$, $\alpha_2\pi$, $\alpha_3\pi$ by 
$$
F(w) = \int_0^w w^{\alpha_1-1}(w-1)^{\alpha_2-1}dw.
$$

\item[2.] 
There are no accessory parameters, as we have already noted.

\item[3.] 
The inverse function $f(z)$ can again be extended to neighboring triangles by reflection over the sides. 

\item[4.] 
This process is particularly interesting when it leads, as in the case of a rectangle, to a meromorphic function.

\item[5.] 
In order that this be so it is necessary that repeated reflections across sides with a common end point should ultimately lead back to the original triangle in an even number of steps.

\end{enumerate}

\end{frame}

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\begin{frame}{2.4. The Triangle Functions of Schwarz. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
 
\item[6.] 
In other words, the angles must be of the form $\pi/n_1, \pi/n_2, \pi/n_3$ with integral denominators.

\item[7.] 
Elementary reasoning shows that the condition
$$
\frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3} = 1
$$
with integral denominators.

\item[8.] 
Elementary reasoning shows that the condition is fulfilled only by the triples $(3,3,3)$, $(2,4,4)$, and $(2,3,6)$. 

\item[9.] 
They correspond to an equilateral triangle, an isosceles right triangle, and half an
equilateral triangle. 

\end{enumerate}

\end{frame}

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\begin{frame}{2.4. The Triangle Functions of Schwarz. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[10.] 
In each case it is easy to verify that the reflected images of the triangle fill out the plane, without overlapping and without gaps.

\item[11.] 
This shows that the mapping functions are indeed restrictions of meromorphic functions, known as the {\color{blue}\it Schwarz triangle functions}. 

\item[12.] 
The reader is urged to draw a picture of the triangle net in each of the three cases. 

\item[13.] 
He will then observe that each triangle function has a pair of periods with nonreal ratio, and is thus an elliptic function.

\item[14.] 
As an exercise, the reader should determine how many triangles there are in a
parallelogram spanned by the periods.

\end{enumerate}

\end{frame}

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\begin{frame}{3.1. Functions with the Mean-value Property. Theorem 6. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
A continuous function $u(z)$ which satisfies condition 
$$
u(z_0) = \frac{1}{2\pi}\int_0^{2\pi} u(z_0+re^{i\theta})d\theta
$$
is necessarily harmonic.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{3.2. Harnack's Principle. Theorem 7. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Consider a sequence of functions $u_n(z)$, each defined and harmonic in a certain region $\Omega_n$.
Let $\Omega$ be a region such that every point in $\Omega$ has a neighborhood  contained in all but a finite number of the $\Omega_n$, and assume moreover that in this neighborhood $u_n(z) \le u_{n+1}(z)$ as soon as $n$ is sufficiently large.
Then there are only two possibilities: either $u_n(z)$ tends uniformly to $+\infty$ on every compact subset of $\Omega$, or $u_n(z)$ tends to a harmonic limit function $u(z)$ in $\Omega$, uniformly on compact sets. 
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{3.2. Harnack's Principle. Exercise 1. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $E$ is a compact aet in a region $\Omega$, prove that there exists a constant $M$, depending only on $E$ and $\Omega$, such that every positive harmonic function
$u(z)$ in $\Omega$ satisfies $u(z_2) \le Mu(z_1)$ for any two points $z_1,z_2 \in E$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{4.1. Subharmonic Functions. Definition 1.}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
What is a convex function? What is a subharmonic function?
}

\item  Answer. 
\begin{enumerate}
\item 
A continuous real-valued function $v(z)$, defined in a region $\Omega$, is said to be subharmonic in $\Omega$ if for any harmonic function $u(z)$ in a region $\Omega'\subset\Omega$ the difference $v-u$ satisfies the maximum principle in $\Omega'$.

\item 
A function $v(x)$ is said to be convex if, in any interval, it is at most equal to the linear function $u(x)$ with the same values as $v(x)$ at the end points of the interval. 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{4.1. Subharmonic Functions. Theorem 8.}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
A continuous function $v(z)$ is subharmonic in $\Omega$ if and only if it satisfies the inequality
$$
v(z_0) \le \frac{1}{2\pi}\int_0^{2\pi} v(z_0+re^{i\theta})d\theta
$$
for every disk $|z-z_0|\le r$ contained in $\Omega$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{4.1. Subharmonic Functions. Exercise - 1}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Show that the functions $|x|$, $|z|^\alpha (\alpha \ge 0)$, $\log(1+|z|^2)$ are subharmonic.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{4.1. Subharmonic Functions. Exercise - 2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $f(z)$ is analytic, prove that $|f(z)|^\alpha (\alpha \ge 0)$ and $\log (1 + |f(z)|^2)$ are subharmonic.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{4.1. Subharmonic Functions. Exercise - 3}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $v$ is continuous together with its partial derivatives up to the second order, prove that $v$ is subharmonic if and only if $\Delta v \ge 0$.
Hint: For the sufficiency, prove first that $v + \varepsilon x^2, \varepsilon > 0$, is subharmonic. For the necessity, show that if $\Delta v < 0$ the mean value over a circle would be a decreasing function of the radius.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{4.1. Subharmonic Functions. Exercise - 4}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove that a subharmonic function remains subharmonic if the independent variable is subjected to a conformal mapping.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{4.1. Subharmonic Functions. Exercise - 5}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Formulate and prove a theorem to the effect that a uniform limit of subharmonic functions is subharmonic.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{4.1. Subharmonic Functions. Exercise - 6}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $v(z)$ is upper semicontinuous on the open set $\Omega$, show that it has a maximum on any compact set $E \subset \Omega$.

}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{4.2. Solution of Dirichlet's Problem. Theorem 9. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
The Dirichlet problem can be solved for any region $\Omega$ such that each boundary point is the end point of a line segment whose other points are exterior to $\Omega$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{4.2. Solution of Dirichlet's Problem. Exercise 1. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $\Omega$ is the punctured disk $0 < |z| < 1$ and if $f$ is given by $f(\zeta)=0$ for $|\zeta| = 1$, $f(0) = 1$, show that all functions $v\in \mathcal{B}(f)$ are $\le 0$ in $\Omega$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{5.1. Harmonic Measures. Theorem 10. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
The function $F(z)$ effects a one-to-one conformal mapping of $\Omega$ onto the annulus $1 < |w| < e^{\lambda_1}$ minus $n-2$ concentric arcs situated on the circles $|w| = e^{\lambda_i}, i = 2,\cdots,n-1$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{5.1. Harmonic Measures. Exercise - 1}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove directly that two circular annuli are conformally equivalent if and only if the ratios of their radii are equal.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{5.1. Harmonic Measures. Exercise - 2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove that $\alpha_{ij} = \alpha_{ji}$. 
Hint: Apply Theorem 21, Chap. 4.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{5.2. Green's Function. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
What is Green's function?
}

\item  Answer. 
\begin{enumerate}
\item 
We consider a point $z_0\in \Omega$ and solve the Dirichlet problem in $\Omega$ with
the boundary values $\log |\zeta - z_0|$. 
The solution is denoted by $G(z)$, but the main interest is attached to the function $g(z) = G(z) - \log |z-z_0|$, known as the Green's function of $\Omega$ with pole at $z_0$. When the dependence on $z_0$ is emphasized, it is denoted by $g(z,z_0)$.

\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{5.3. Parallel Slit Regions. Theorem 11. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
The mappings determined by $p(z)$ and $q(z)$ are one to one, and the image of $\Omega$ is a slit region whose complement consists of $n$ vertical or horizontal segments, respectively (Fig. 6-5a, b).
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{5.3. Parallel Slit Regions. Exercise - 1 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove that $g(z,z_0)$ is simultaneom;ly continuous in both variables, for $z \neq z_0$. Hint: Apply the maximum-minimum principle to $G(z,z_0)$. 
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{5.3. Parallel Slit Regions. Exercise - 2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Show that the function $e^{-i\alpha}(q\cos\alpha + ip\sin\alpha)$ maps $\Omega$ onto a region bounded by inclined slits.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{5.3. Parallel Slit Regions. Exercise - 3}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Using Ex. 2, show that $p + q$ maps $\Omega$ in a one-to-one manner onto a region bounded by convex contours.

}

\item  Comments:
\begin{enumerate}%\itemsep1.8cm
\item  %(i) 
A closed curve is said to be convex if it intersects every straight line at most twice.

\item  %(ii) 
To prove that the image of $C_k$ under $p + q$ is convex we need only show that for every $\alpha$ the function $\mathrm{Re} (p + q)e^{i\alpha}$ takes no value more than twice on $C_k$.
But $\mathrm{Re} (p + q)e^{i\alpha}$ differs from $\mathrm{Re} (q\cos\alpha + ip\sin\alpha)$ only by a constant, and the desired conclusion follows by the properties of the mapping function in Ex. 2.

\item  %(iii) 
Finally, the argument principle can be used to show that the images of the contours $C_k$ have winding number 0 with respect to all values of $p + q$. This implies, in particular, that the convex curves lie outside of each other.

\end{enumerate}

\end{itemize}

\end{frame}

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